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(B) The potential $V$ at the surface of a spherical conductor of radius $R$ with charge $Q$ is given by $V = \frac{Q}{4 \pi \varepsilon_0 R}$.Let $Q_1

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$\frac{R_2}{R_1}$

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(B) The potential $V$ at the surface of a spherical conductor of radius $R$ with charge $Q$ is given by $V = \frac{Q}{4 \pi \varepsilon_0 R}$.Let $Q_1$ and $Q_2$ be the charges on the two spheres with radii $R_1$ and $R_2$ respectively.When the two spheres are connected by a wire,charge flows until their potentials become equal,i.e.,$V_1 = V_2$.Since surface charge density $\sigma = \frac{Q}{4 \pi R^2}$,we have $Q = \sigma \cdot 4 \pi R^2$.Substituting this into the potential formula: $V = \frac{\sigma \cdot 4 \pi R^2}{4 \pi \varepsilon_0 R} = \frac{\sigma R}{\varepsilon_0}$.Equating the potentials: $\frac{\sigma_1 R_1}{\varepsilon_0} = \frac{\sigma_2 R_2}{\varepsilon_0}$.Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.

Two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a wire. The ratio of surface charge densities of the spheres $\sigma_1/\sigma_2$ will be

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